What is the equation in standard form of the parabola with a focus at (4,3) and a directrix of y= -3?

Jul 11, 2017

$y = \frac{1}{12} {x}^{2} - \frac{2}{3} x + \frac{4}{3}$

Explanation:

The focus must be the same distance from the vertex as the directrix for this to work. So apply the Midpoint theorem:$M = \left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$
$\setminus \therefore \left(\frac{4 + 4}{2} , \frac{3 + \left(- 3\right)}{2}\right)$ (both have the same x-value for convenience)
which gets you a vertex of $\left(4 , 0\right)$. This means both focus and directrix are 3 vertical units away from the vertex ($p = 3$).

Your vertex is the coordinate $\left(h , k\right)$, so we input into the vertical parabola format...

$4 \left(3\right) \left(y - 0\right) = {\left(x - 4\right)}^{2}$
$12 \left(y - 0\right) = {\left(x - 4\right)}^{2}$

Now we simplify.
$12 y - 0 = \left(x - 4\right) \left(x - 4\right)$
$12 y = {x}^{2} - 8 x + 16$

Standard form is $y = a {x}^{2} + b x + c$ but we have to isolate the $y$ on the left. So divide everything by 12 and you have your answer.

$y = \frac{1}{12} {x}^{2} - \frac{8}{12} x + \frac{16}{12}$
$y = \frac{1}{12} {x}^{2} - \frac{2}{3} x + \frac{4}{3}$